The Baudhayana-Pythagoras Theorem · Lesson 7 of 9
Further Applications of the Baudhāyana–Pythagoras Theorem
“Find hidden right triangles in rectangles, rhombi, equilateral triangles, and the lotus problem.”
• Draw the right triangle needed to model a geometric problem. • Use half-diagonals to calculate the side of a rhombus. • Find an equilateral triangle’s height and area. • Translate the lotus situation into an equation and solve it. • State assumptions and check results in the original situation.
A problem does not always arrive as a triangle with two neatly labelled sides. It might ask for the diagonal of a screen, the side of a rhombus, or the depth of water beneath a lotus flower. The right triangle may be hidden inside the situation. Drawing the right extra line is often more important than doing the final arithmetic.
Your task is to turn each situation into a diagram that preserves the given information. Mark known lengths, introduce a letter for the unknown, and identify the right angle. Then connect the diagram to the theorem rather than choosing numbers just because they appear in the question.
We will move from familiar shapes to a situation involving an unknown length in two different expressions. The same area relationship will work throughout, but each problem needs its own geometric explanation.
Finding the Diagonal of a Square or Rectangle
A diagonal joins opposite vertices. In a rectangle, it divides the shape into two right triangles. Each triangle has the rectangle’s length and width as perpendicular sides, so the diagonal is the hypotenuse.
For a square, the two perpendicular sides are equal. If its side is s, then d² = s² + s² = 2s² and d = s√2. For a rectangle with different length and width, keep the two squared terms separate.
Problem
Find the diagonal of a square with side 5 cm.
- 1.The diagonal is opposite the right angle in a triangle with perpendicular sides 5 and 5.
- 2.Thus d² = 25 + 25 = 50.
- 3.The exact diagonal is √50 = 5√2 cm.
- 4.Because 7.0² = 49 and 7.1² = 50.41, its length lies between 7.0 and 7.1 cm.
Problem
A rectangle has sides 8 cm and 15 cm. Find its diagonal.
- 1.A diagonal forms a right triangle with perpendicular sides 8 and 15.
- 2.Calculate d² = 8² + 15² = 64 + 225 = 289.
- 3.Hence d = 17 cm.
- 4.The diagonal is longer than either side, as a hypotenuse should be.
Use five known triples to give rectangle side pairs and their diagonals: 3 and 4 give diagonal 5; 5 and 12 give 13; 8 and 15 give 17; 7 and 24 give 25; 12 and 35 give 37. Check each by adding the squares of the side lengths. A rectangle has two equal diagonals, so the same calculated length applies to both.
Finding a Rhombus’s Side from Its Diagonals
A rhombus has four equal sides, but its angles need not be right angles. Applying the theorem directly to two adjacent sides of a general rhombus would therefore be unjustified. Its diagonals provide the needed right triangles.
The diagonals of a rhombus bisect one another at right angles. Their intersection divides each full diagonal into two equal parts. A side of the rhombus is the hypotenuse of a triangle whose perpendicular sides are these half-diagonals.
Consequently, first halve both diagonal lengths. Square those halves and add them. Using the full diagonals would calculate a triangle twice as large in every length, not the triangle inside the given rhombus.
Problem
Find the side length of the rhombus.
- 1.Half the first diagonal is 24 ÷ 2 = 12 units.
- 2.Half the second diagonal is 70 ÷ 2 = 35 units.
- 3.The side s satisfies s² = 12² + 35² = 144 + 1225 = 1369.
- 4.Therefore s = 37 units. Each of the four rhombus sides has this length.
Problem
Find the side of a rhombus whose diagonals are 10 cm and 24 cm.
- 1.The perpendicular half-diagonals measure 5 cm and 12 cm.
- 2.A side has square 5² + 12² = 25 + 144 = 169.
- 3.Thus each side is 13 cm.
- 4.Using 10 and 24 directly would give 26 cm, which is twice the correct side length.
Finding the Height of an Equilateral Triangle
In an equilateral triangle, all three sides have the same length. Draw a perpendicular from the top vertex to the opposite side. This perpendicular is an altitude, meaning a height measured at right angles to the chosen base.
The two resulting right triangles have equal hypotenuses because the original sloping sides are equal. They also share the altitude as a common side. By right-angle–hypotenuse–side congruence, the triangles are congruent, so their base segments are equal. Thus the altitude bisects the base.
If the original side is 6 units, each half-base is 3 units. In either small triangle, the hypotenuse is 6, not the altitude. Therefore the altitude’s square is 6² − 3². This is a subtraction problem because the hypotenuse is known.
Problem
Find the area of an equilateral triangle with side 6 units.
- 1.Draw its altitude. The base is bisected into lengths 3 and 3.
- 2.In either right triangle, h² + 3² = 6², so h² = 36 − 9 = 27.
- 3.The height is √27 = 3√3 units.
- 4.Use area = ½ × base × height = ½ × 6 × 3√3.
- 5.The exact area is 9√3 square units. Keep length units for h and square units for area.
Problem
Find the height and area of an equilateral triangle of side 4 cm.
- 1.Its half-base is 2 cm, so h² = 4² − 2² = 16 − 4 = 12.
- 2.Hence h = √12 = 2√3 cm.
- 3.Area = ½ × 4 × 2√3 = 4√3 cm².
- 4.The method is unchanged: justify the half-base, find height, then calculate area.
A Problem from Bhāskarāchārya’s Līlāvatī
A lotus flower stands above the surface of a lake. Its tip is 1 unit above the water. When it is moved sideways, the tip reaches the water surface at a point 3 units from its original vertical position. How deep is the water at the point where the stem is rooted?
Let x be the water depth. Initially, x units of stem are below the water and 1 unit is above it, so the total stem length is x + 1. In the second position, that same length becomes the sloping side of a triangle.
For this idealised model, assume the original stem is vertical, its root remains fixed, its length stays unchanged, and it is represented by a straight segment in each position. The vertical depth and the horizontal displacement then meet at a right angle. The triangle’s perpendicular sides are x and 3, while its hypotenuse is x + 1.
Problem
Find the water depth for a stem extending 1 unit above water and reaching the surface 3 units away.
- 1.Write x² + 3² = (x + 1)².
- 2.Expand the square: x² + 9 = x² + 2x + 1. The term 2x must not be omitted.
- 3.Subtract x² from both sides: 9 = 2x + 1.
- 4.Subtract 1 to get 8 = 2x, then divide by 2: x = 4.
- 5.The water depth is 4 units and the stem length is 5 units.
- 6.Check in the situation: 4² + 3² = 16 + 9 = 25 = 5², and a 5-unit stem extends 1 unit above 4-unit-deep water.
At first, the data seemed insufficient because no complete stem length was supplied. The useful observation was that its length could be expressed in terms of the unknown depth. The theorem then related these quantities, and the x² terms cancelled.
A statement such as “the stem is x + 1” must come from the situation, not from a memorised formula. If the flower were 2 units above the water, the stem length would instead be x + 2 and the expansion would change accordingly.
Turning a Situation into a Mathematical Model
Use a consistent routine: sketch the situation, mark the right angle, label the known distances, and introduce one clear unknown. Decide whether a labelled distance is a full length or half a length. Then write the equation before calculating.
After solving, return to the original question. A height is not automatically an area, a half-diagonal is not a full diagonal, and the stem length is not the water depth. The same numerical work can answer the wrong question if you forget what the variable means.
| Right-triangle data | Equation for the missing length | Result |
|---|---|---|
| Perpendicular sides 7 and 9 | c² = 49 + 81 = 130 | c = √130 |
| Perpendicular sides 4 and 10 | c² = 16 + 100 = 116 | c = √116 |
| Side 40; hypotenuse 41 | b² = 1681 − 1600 = 81 | b = 9 |
| Side 10; hypotenuse √200 | b² = 200 − 100 = 100 | b = 10 |
| Perpendicular sides 10 and √150 | c² = 100 + 150 = 250 | c = √250 |
| Side 27; hypotenuse 45 | b² = 2025 − 729 = 1296 | b = 36 |
The table includes six cases whose drawings can be rotated in any direction. A radical label such as √200 is a length, so its square is 200. Do not read √200 as 200. The correct equation follows the right-angle mark, not the orientation of the picture.
Quiz
Which sides form the right triangle used for a rectangle’s diagonal?
A rhombus has diagonals 16 and 30. What is its side length?
For an equilateral triangle of side 10, what half-base is used to find its altitude?
In the lotus problem, why is the sloping side x + 1?
If the perpendicular sides are 10 and √150, what is the hypotenuse squared?
Practice Problems
- Find the diagonal of a rectangle with sides 12 cm and 16 cm. Solution: 1. d² = 144 + 256 = 400. 2. Therefore d = 20 cm.
- A rhombus has diagonals 18 cm and 24 cm. Find its side and perimeter. Solution: 1. The half-diagonals are 9 cm and 12 cm. 2. s² = 81 + 144 = 225, so s = 15 cm. 3. All four sides are equal, so the perimeter is 4 × 15 = 60 cm.
- Find the height and area of an equilateral triangle of side 8 units. Solution: 1. The altitude bisects the base into two 4-unit segments. 2. h² = 64 − 16 = 48, so h = 4√3 units. 3. Area = ½ × 8 × 4√3 = 16√3 square units.
- A stem extends 2 units above water and reaches the surface 6 units away. Under the straight-stem model, find the depth. Solution: 1. Let depth be x, giving unchanged stem length x + 2. 2. Use x² + 36 = (x + 2)² = x² + 4x + 4. 3. Cancel x²: 36 = 4x + 4, so 32 = 4x and x = 8. 4. The stem is 10 units long; 8² + 6² = 100 confirms the result.
- A student uses full diagonals 24 and 70 to find a rhombus side as 74. Explain the error and find the correct side. Solution: 1. The right triangle inside the rhombus uses half-diagonals, not full diagonals. 2. Those lengths are 12 and 35. 3. The correct side is √(144 + 1225) = √1369 = 37 units. 4. The incorrect triangle was twice as large in every length, so its 74-unit hypotenuse was twice the required answer.
Key Takeaways
• A useful diagram reveals the right triangle hidden inside a problem. • Rectangle diagonals are hypotenuses; square diagonals equal side × √2. • A rhombus side uses the perpendicular half-diagonals. • An equilateral triangle’s altitude bisects its base; use the altitude to find area. • The lotus problem relates depth x to unchanged stem length x + 1. • State modelling assumptions and check the result in the original situation.